Problem 83 - College Physics

Question

\(\bullet\) Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortu- nately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

Step-by-Step Solution

Answer

The average power supplied is 390.0 W.

Step 1: Calculate the acceleration

To find the acceleration, we need to use the formula for velocity with constant acceleration: \[ v = u + at \] Where \( u = 0 \) (initial velocity), \( v = 6.00 \, \text{m/s} \) (final velocity), and \( t = 3.00 \, \text{s} \). Solving for \( a \): \[ 6.00 \, \text{m/s} = 0 + a \times 3.00 \, \text{s} \] \[ a = \frac{6.00 \, \text{m/s}}{3.00 \, \text{s}} = 2.00 \, \text{m/s}^2 \]

Step 2: Determine the force applied

Using Newton's second law, we determine the force applied with: \[ F = m \times a \] Where \( m = 65.0 \, \text{kg} \) (mass of your friend) and \( a = 2.00 \, \text{m/s}^2 \) (acceleration). Substitute the values: \[ F = 65.0 \, \text{kg} \times 2.00 \, \text{m/s}^2 = 130.0 \, \text{N} \]

Step 3: Calculate work done

Work done \( W \) by the force is given by the product of force and displacement: \[ W = F \times s \] First, find the displacement \( s \) using \( v^2 = u^2 + 2as \):\[ 6.00^2 = 0 + 2 \times 2.00 \times s \] \[ 36 = 4s \] \[ s = \frac{36}{4} = 9.00 \, \text{m} \] Then calculate work: \[ W = 130.0 \, \text{N} \times 9.00 \, \text{m} = 1170.0 \, \text{J} \]

Step 4: Determine average power

Power is the rate at which work is done, given by: \[ P = \frac{W}{t} \] Where \( W = 1170.0 \, \text{J} \) (work done) and \( t = 3.00 \, \text{s} \). Substitute the values: \[ P = \frac{1170.0 \, \text{J}}{3.00 \, \text{s}} = 390.0 \, \text{W} \]

Key Concepts

Newton's Second Law

Newton's Second Law is fundamental for understanding how forces influence the motion of objects. At its core, it describes the relationship between an object's mass, its acceleration, and the applied force through the formula: \[ F = m \times a \] Here, \( F \) denotes the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration it experiences. This equation tells us that the force applied on an object is directly proportional to the acceleration produced, provided the mass stays constant. In simpler terms, if you increase the mass, you'll need more force to get the same acceleration. Or, for a given force, a heavier object will have a smaller acceleration. Understanding this principle is crucial, as it sets the foundation for analyzing dynamic systems, such as moving vehicles, dropped objects, or in our exercise case, a person being pulled across ice.

Acceleration Calculation

Acceleration is a measure of how quickly an object's velocity changes. In physics, it's important to understand that acceleration is not just about speed, but also about direction. It's calculated using the formula from constant acceleration equations: \[ v = u + a \times t \] Where: - \( v \) is the final velocity.
- \( u \) is the initial velocity.
- \( a \) is the acceleration.
- \( t \) is the time over which the acceleration occurs. In our exercise example, the friend starts from rest, meaning the initial velocity \( u \) is zero. They reach a speed of 6.00 m/s over 3 seconds. The calculation showed an acceleration of 2.00 m/s², which was derived by rearranging the formula to solve for \( a \). This comprehension is vital for predicting how quickly an object can change its velocity, influencing many practical scenarios in science and engineering.

Work and Energy

Work in physics is a measure of energy transfer that happens when a force moves an object over a distance. The fundamental equation for calculating work is:\[ W = F \times s \] Where: - \( W \) is the work done.
- \( F \) is the force applied.
- \( s \) is the displacement of the object in the direction of its force. In the context of our exercise, once the force was determined (130.0 N), the next step was to find how far the object had traveled. The distance \( s \) was calculated using kinematic equations and found to be 9.00 m. Therefore, the work done to move the friend across the ice was 1170.0 Joules. Understanding the concept of work helps illustrate how energy is transferred through forces, essential for analyzing any system where forces are involved.

Average Power Calculation

Power is essentially the rate of doing work - or how fast energy is being transferred or transformed. In the exercise, average power is calculated from the work done over a certain period. The formula to use here is:\[ P = \frac{W}{t} \] Where: - \( P \) is the average power.
- \( W \) is the work done.
- \( t \) is the time over which the work is done. By plugging in the values from our practical scenario: 1170.0 J of work over 3.00 s, we found the average power supplied to be 390.0 Watts. This concept is particularly significant in fields like engineering and physics, where understanding the efficiency and effectiveness of systems is key. By knowing the power, you can assess how much energy is used or required to perform a task over time.